# F.5 Maths Module 2 question

Consider the curve C: x^(2) + 4y^(2) = 68. Find the equation of the normals to C that pass through (0,3)

Thank you :)

### 3 個解答

• 最佳解答

x^2 + 4y^2 = 68

Differentiate both sides w.r.t. x, we get

2x + 8y dy/dx = 0

==> dy/dx = -x/(4y)

So, the slope of the normal at any point on this curve is 4y/x.

Suppose (x, y) is the required point on this curve, then

(y - 3)/(x - 0) = 4y/x

==> x(y - 3) = 4xy

==> x(y - 3) - 4xy = 0

==> x(y - 3 - 4y) = 0

==> -3x(y + 1) = 0

==> x = 0 or y = -1

When x = 0,

4y^2 = 68

==> y = √17 or y = -√17When y = -1,x^2 + 4(-1)^2 = 68==> x = 8 or -8equation of normal passes thru' (0, 3) and (0, √17) is x = 0;equation of normal passes thru' (0, 3) and (0, -√17) is x = 0;equation of normal passes thru' (0, 3) and (8, -1) is :(y - 3)/(x - 0) = 4*(-1)/8==> x + 2y - 6 = 0equation of normal passes thru' (0, 3) and (-8, -1) is :(y - 3)/(x - 0) = 4*(-1)/(-8)==> x - 2y + 6 = 0

Therefore, the equation of normals are :(i) x = 0(ii) x + 2y - 6 = 0(iii) x - 2y + 6 = 0

(Totally there are 3 normals pass thru' (0, 3))

• x^2+〖4y〗^2=68

2x+4(2y)dy/dx=0

dy/dx=(-2x)/8y

dy/dx=(-x)/4y

When x=0, y=3

dy/dx=(-0)/3=0

Equation of the normal to C:

y-3=(-1)/0(x-0)

y=3

• to get the equation of tangent.

The ellipse is :

x²/(√68)² + y²/(√17)² = 1

Tangent at (0,3) is 0x/(√68)² + 3y/(√17)² = 1

3y = 17

slope of tangent = 0

slope of normal is -∞

equation of normal at (0,3) is x=0;